How do you find the exact relative maximum and minimum of the polynomial function of #f(x)=x^3-3x+6#?

#f(x) = x^3-3x+6#

Find critical numbers for #f# .

#f'(x) = 3x^2-3# never fails to exist and is #0# at the solutions of

#3x^2-3=0#.

The solution are #+-1#, so the critical numbers are #+-1#.

Test the critical numbers

Use either the first or second derivative test to see that

#f(-1)# is a relative maximum.

(Second derivative test: #f''(x) = 6x#, so #f''(-1) < 0# and we conclude #f(-10# is a relative maximum.)

And #f(1)# is a relative minimum.

(f''(1) > 0#, so #f(1) is a relative minimum.)

Find the minimum and maximum values

Rel max: #f(-1)=(-1)^3-3(-1)+6 = 8#

Rel min: #f(1) = (1)^3-3(1)+6 = 4#